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Regular expression (sed)

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# 1  
Old 10-19-2010
Regular expression (sed)
Hi

I need to get text that are within ""

For example

File:
asdasd "test test2" sadasds asdda asdasd "demo demo2"

Output:
test test2 demo demo2

Any help is good

Thank you
# 2  
Old 10-19-2010
Code:
$ sed 's/\(.*\)\(\".*\"\)\(.*\)\(\".*\"\)/\2 \4/g;s/\"//g' file
test test2 demo demo2

This User Gave Thanks to cabrao For This Post:
blito_loco
# 3  
Old 10-19-2010
Code:
sed 's/[^"]*"\([^"]*\)"[^"]*/\1 /g' infile

This User Gave Thanks to Scrutinizer For This Post:
blito_loco
# 4  
Old 10-19-2010
Quote:
Originally Posted by cabrao
Code:
$ sed 's/\(.*\)\(\".*\"\)\(.*\)\(\".*\"\)/\2 \4/g;s/\"//g' file
test test2 demo demo2

I works

Code:
echo '--yyy "test test2" sadasds --xxx "demo demo2"' | sed 's/\(.*\)\(\".*\"\)\(.*\)\(\".*\"\)/\2 \4/g;s/\"//g'

Output: test test2 demo demo2

But, now i need obtain only text that start with --xxx

Output:
demo demo2

Thanks for your quick help
# 5  
Old 10-19-2010
Quote:
Originally Posted by blito_loco
asdasd "test test2" sadasds asdda asdasd "demo demo2"
Code:
$ ruby -ne 'puts $_.scan(/\".[^"]*?\"/)' file
"test test2"
"demo demo2"

Code:
$ echo '--yyy "test test2" sadasds --xxx "demo demo2"' | ruby -e 'print gets.split("--xxx")[-1]'
 "demo demo2"

This User Gave Thanks to kurumi For This Post:
blito_loco
# 6  
Old 10-19-2010
Assuming one --xxx per line, and that there might be text following the quoted string, this will work:

Code:
 sed -r 's/.*--xxx[^"]*"//; s/".*//'

Last edited by agama; 10-19-2010 at 11:34 PM.. Reason: typo
This User Gave Thanks to agama For This Post:
blito_loco
# 7  
Old 10-20-2010
Code:
# echo '--yyy "test test2" sadasds --xxx "demo demo2"' | sed 's/.*-xxx \(.*\)/\1/'
"demo demo2"

Code:
# echo '--yyy "test test2" sadasds --xxx "demo demo2"' | sed 's/.*-xxx "\(.*\)"/\1/'
demo demo2

This User Gave Thanks to ygemici For This Post:
blito_loco
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